-
1
1-2
−
−
−
−
−
− − −
≥ −
≥−
For Chapter 1
∑( ) ∑( )
∑( ) ∑( )
∑ ( )
=0 =0
=0 =0
=0even
1
1
1
n
k
n
k
n k k n n
n
k
k
n
k
n k k n
n
kk
n
n n
n
n
k
n
k
n >
n
k
n
k.
n
k.
n / / n
n n
n
n
n
n
n n
n
n
Solutions, answers, and hints for selected problems
Asterisks in “A Modern Approach to Probability Theory” by
Fristedt and Gray
identify the problems that are treated in this supplement. For
many of those
problems, complete solutions are given. For the remaining ones,
we give hints,
partial solutions, or numerical answers only.
. Method 1: By the Binomial Theorem,
= 1 1 = (1 + 1) = 2
and, for 0,
( 1) = 1 ( 1) = (1 1) = 0
Addition and then division by 2 gives
= 2
The answer for positive is 2 2 = 1 2. The answer for = 0 is
easily seen to
equal 1.
Method 2: For 1 consider a sequence of length ( 1). If it
contains an even
number of ‘heads’, adjoin a ‘tails’ to it to obtain a length-
sequence containing an even
number of ‘heads’. If it contains an odd number of ‘heads’,
adjoin a ‘heads’ to it to
obtain a length- sequence containing an even number of ‘heads’.
Moreover, all length-
sequences containing an even number of ‘heads’ are obtained by
one of the preceding
two procedures. We have thus established, for 1, a one-to-one
correspondence
between the set of all length-( 1) sequences and the set of
those length- sequences
that contain an even number of ‘heads’. Therefore, there are 2
length- sequences
that contain an even number of ‘heads’. To treat the remaining
case = 0, we observe
-
⋃ ⋃
∑ ∑
−
− −
∈ ∈
∈ ∈
136
23
116
35
3
78
58
1-4
1-10
1-11
1-12
≤ ≤ ≤ ≤
→∞
∈
E
E
∪ − ∩
∩
∩ ∩
≤ ≤ − −
∩ ≤→∞ ∩
j
n n
s
s
s s s s
s S s s S s
s S
s
s S
s
c c
c c
c c c ck
k k
kk
c ck
k
c c
2 SOLUTIONS, ANSWERS, AND HINTS FOR SELECTED PROBLEMS
r, g r g r g
A
A n A
n
j
S
s
, , , , s
s s S A
ω s , , , ,
, , B ω
s , , , ,
A B P A P B
A A B B
A ω , , , , B
ω , , , , A B
P A P A P B P B .
A B
P A P B P A B P A B .
P A B
P A P B P A
P A B A B D
ω k
j k , , j j j
k ω
D
, , , , P A B P D
k P A B
that the empty sequence, which is the only length-0 sequence,
contains zero ‘heads’.
Since 0 is even, there is 1 length-0 sequence containing an even
number of ‘heads’.
. 2
. The thirty-six points to each of which is assigned probability
are the ordered
pairs ( ) for 1 6 and 1 6. The coordinates and represent the
numbers
showing on the red die and green die, respectively.
. The set consisting of a single sample point, being the
intersection of countably
many events of the form (1.2), is an event. Its probability is
no larger than that of
any such . For each and each sample point, there is such an that
has probability
2 . Thus, the probability of that sample point is no larger than
2 . Letting
we see that the probability of the sample point is 0. The
process of flipping a coin until
the first tails occurs terminates in a finite number of steps
with probability 1.
. (i) Sum the answer to Problem 4 over odd positive to obtain
.
(ii)
(iii) (Caution: it is common for students to use invalid
reasoning in this type of
problem.) We use ‘1’ and ‘0’ to denote heads and tails,
respectively. Let denote the
set of finite sequences of 1’s and 0’s terminating with 1,
containing no subsequence of
the form (1 0 1) or (1 1 1), and having the additional property
that if the length of
is at least two, then the penultimate term in is 0. For each ,
let be the event
consisting of those infinite sequences that begin with followed
by (0 1 1), (0 1 0),
or (1 0 1) in the next three positions, and let be the event
consisting of those
that begin with followed by (1 1 1) or (1 1 0) in the next three
positions. Note that
each and is a member of . Clearly 2 ( ) = 3 ( ).
Let = and = . Straightforward set-theoretic arguments show
that consists of those in which (1 0 1) occurs before (1 1 1),
consists of those
in which (1 1 1) occurs before (1 0 1). By writing and as
countable unions
of members of , we have shown that they are events. Note that in
each case, these
unions are taken over a family of pairwise disjoint events, from
which it follows that
2 ( ) = 2 ( ) = 3 ( ) = 3 ( )
Also, and are clearly disjoint, so
( ) + ( ) = ( ) = 1 ( )
We will show that ( ) = 0, so that the above two equalities
become two
equations in the two unknowns ( ) and ( ), the solution of which
gives ( ) = .
To show that ( ) = 0 we note that is a subset of the event
consisting of those that begin with a sequence of length 3
having the property that,
for 1 , the sequence (1 1 1) does not occur in positions 3 2, 3
1, 3 . The
number of ways of filling the first 3 positions of with 1’s and
0’s is 2 = 8 . The
number of ways of doing it so as to obtain a member of is 7 (7
choices for positions
1 2 3; 7 choices for positions 4 5 6 and so forth.). Thus, ( ) (
) = ( ) .
Now let to obtain the desired conclusion, ( ) = 0.
(iv)
-
−
c
c
c
Hint:
For Chapter 2
1-14
1-16
2-2
2-3
+
+
+
+ +
+ +
+
+ +
+
+
+
+
+
+
1
R R
R
R
R R
R R
R R
R R R
R
R
R
R
R
R
R
R
SOLUTIONS, ANSWERS, AND HINTS FOR SELECTED PROBLEMS 3
B C
G { ∈ B ⊆ }
C GG
B B GG ∩
B B G
∩B G
GC
G ⊆ C D −∞
H { ∪ ∈ C ∈ D}
B ⊆ H CH H
∪ \ ∪ −∞ \⊆ ⊆ −∞ H
∈
∩ ∪ −∞ ∩
−∞ HB
E
{ ∈ } { ∈ }
{ ∈ }{ ∈ } { ∈ 6 }{ ∈ }
σ σ
B B .
σ
G G
G
σ
O O
σ σ
σ ,
C D C , D .
C D C , D
C D ,
σ O
O O , O
O
, σ σ
X B X X B
X
X
B X Y
P ω X ω B P ω Y ω B .
P ω X ω B
P ω X ω B, Y ω X ω P ω X ω B, Y ω X ω
P ω X ω Y ω B, Y ω X ω .
Let denote the Borel -field of , the Borel -field of , and
= :
The goal is to prove = .
We first prove that is a -field of subsets of . Countable unions
of members of
are members of and unions of subsets of are subsets of . Hence,
is closed
under countable unions. The complement in of a member of equals
,
where denotes the complement in . This set is clearly a subset
of and it is also
a member of because it is the intersection of two members of .
Therefore, is a
-field.
The open subsets of have the form , where is open in . Such
sets, being
subsets of and intersections of two members of , are members of
. Thus, the
-field contains the -field generated by the collection of these
open subsets—namely
.
To show that we introduce the Borel -field of subsets of ( 0)
with the
relative topology and set
= :
We can finish the proof by showing that , because consists of
those members
of which are subsets of . It is clear that is closed under
countable unions. The
formula
( ) = ( ) (( 0) )
for and ( 0) shows that it is closed under complementation. So
is
a -field. For any open set , the representation
= ( ) (( 0) )
represents as the union of open, and therefore Borel, subsets of
the spaces and
( 0). Thus, the -field contains the -field generated by the
collection of open
subsets of —namely .
It suffices to show that every open set is the union of open
boxes having
edges of rational length and centers with rational
coordinates.
. Let be a continuous function. For any open of the target of ,
( ) is
open by continuity, and thus is an event in the domain of . Now
apply Proposition 3
with equal to the collection of open subsets in the target of
.
. Let be an arbitrary measurable set in the common target of and
. We need
to show that
( : ( ) ) = ( : ( ) )
Here is the relevant calculation:
( : ( ) )
= ( : ( ) ( ) = ( ) ) + ( : ( ) ( ) = ( ) )
= ( : ( ) and ( ) ( ) = ( ) )
-
R
( )k
πππ
Hint:
−
√
√
−√
−
For Chapter 3
2-9
2-12
2-14
2-19
2-21
2-22
3-3
1
2 1 2
1 2
1
59
1 2 1 2
12
2
1+ 22
12
2+ 22
2 22
116
28
def
{ ≤ }
−∞∈
≤ ≤
∈ −∞−∞
{ }
{ }{ ≥ } { ≥ }
{ } { ≥ } ≤
∈ ≥
− ≥−
− ≥ ≥ −
↘
4 SOLUTIONS, ANSWERS, AND HINTS FOR SELECTED PROBLEMS
X Y
A ω X ω c c c
a A
, a A ω < a a A
ω A ω < ω
X ω X ω c ,
ω A A , a
, a
v , v < v < v < B
B
X
X
k
k/
k/.
ω F x F x < ω
x F x ω x F x ω Y ω
x F x < ω x F x ω Y ω X ω
Y ω X ω Y ω < X ω
x Y ω ,X ω F x ω
X ω F x < ω Y ω Y X
Y
Y Y ω
Y ω ω δ > u > Y ω δ F u < ω
τ < ω F u < τ
Y ω Y τ u > Y ω δ .
δ
In this calculation, we used the fact that the event in the
second term of the second
line is contained in a null event. To complete the proof, carry
out a similar calculation
with the roles of and reversed.
. By Problem 13 of Chapter 1 and Proposition 3 we only need show
that the set
= : ( ) is a Borel set for every (or even just for every
rational ). Let
equal the least upper bound of . We will prove that every member
of the interval
( ) belongs to . Suppose . Since is the least upper bound of ,
there
exists for which . Then
( ) ( )
from which it follows that . Thus, is an interval of the form (
) or
( ] and is, therefore, Borel.
.
. The distribution is uniform on the triangle ( ) : 0 1 . If
is a set for which area is defined, the value that the
distribution assigns to is twice
its area, the factor of 2 arising because the triangle has area
. To prove that is a
random variable— Prove that is continuous, or, alternatively,
avoid the issue
of continuity of a -valued function by first doing Problem 16
and then using it in
conjunction with a proof that each coordinate function is
continuous.
. In case is divisible by 4, the answer is
2
42
Otherwise, the answer is 0.
. The Hausdorff distances are between the first two; between the
first and
third; between the second and third.
. These are the probabilities: , , .
. Fix . Since is increasing, every member of : ( ) is less than
every
member of : ( ) and is thus a lower bound of : ( ) . Hence ( )
=
sup : ( ) is a lower bound of : ( ) . Therefore ( ) ( ).
To prove ( ) = ( ), suppose, for a proof by contradiction, that
( ) ( ),
and consider an ( ( ) ( )). Either ( ) contradicting the
defining property
of ( ) or ( ) contradicting the defining property of ( ). Thus =
, and
we will work with in the next paragraph.
Clearly, is increasing. Thus, to show left continuity we only
need show ( )
( ) for every . Let 0. There exists ( ) for which ( ) .
Hence
there exists such that ( ) . Therefore
( ) ( ) ( )
Now let 0.
-
2
2 2
∫
( ) ( )
∫ ∣∣
−
−
∞− − ∞
− −
−
√
b
a
j
j
π xa
π a x b
b a
ax ax
a a
x
3-8
3-12
3-23
3-28
3-30
3-33
−∞ ∞
{ ≤ } { ≤ − } −
−− − −
− −
− − − − − − − −
−
{ ≤ ≤ } −
√−√
≤
2
1 2
1 2
2 1
2 1 1 1
1 2
2 1
1 1
1 2
2 1
1(1+ )
( +( ) )
1
00
2 3
1
12
SOLUTIONS, ANSWERS, AND HINTS FOR SELECTED PROBLEMS 5
a b
Q a, bπ x
dx .
a b Q a, b
Q a, b Q a, b a > b <
Q a, b a b
F F
X X F
X F aX b
a b a >
F x P ω aX ω b x P ω X ω x b /a F x b /a .
F F
F F a b a >
X F
aX b F
F F
X b F F x
b F b x x
F xx
π
x
πx
π
x
πF x .
X
x a b aX b x
a, b
a, b
ae dx e
P ω X ω e e
a
g x f x f x x > g x x
. Whether or is finite or infinite,
(( )) =1
(1 + )
When and are finite this formula is also a formula for ([ ]),
and similarly for
([ )) and (( ]) in case or , respectively. Note that the
formula
for ([ ]) is correct in the special case = .
. Explanation for ‘type’ only. Suppose first that and are of the
same type.
Then there exist random variables and of the same type such that
is the
distribution function of . Then is also the distribution
function of + for
some and with 0. Thus
( ) = ( : ( ) + ) = ( : ( ) ( ) ) = (( ) )
That is and must satisfy (3.2).
Conversely, suppose that and satisfy (3.2) for some and with 0.
Let
be a random variable with distribution function . the above
calculation then
shows that + is a random variable whose distribution function is
. Therefore
is of the same type as .
. is symmetric about if and only if its distribution function
satisfies (
) = 1 (( ) ) for all .
For the standard Cauchy distribution
( ) =1
2+
arctan=
1
2+
arctan( )
= 11
2+
arctan( )= 1
1
2+
arctan(( ) )= 1 (( ) )
. A random variable having the Cauchy distribution of Problem 8
has density
. For positive and real the continuous density of + is
.
The density of the uniform distribution with support [ ] is on
the interval
[ ] and 0 elsewhere.
.
= = 1
( : 2 ( ) 3 ) =
median = log 2
. ( ) = [ ( ) + ( )] if 0 and ( ) = 0 if 0.
-
2
2
2
Hint:
For Chapter 4
3-34
3-40
3-41
∫ ∫∫
∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
√
−
−
−
√
√
−
−
−
≥
{ − ≤ } { ≥ −
≥ ≤
− −
∞−
∞−
∞− −
∞− −
∞−
∞ ∞− − −
∞ ∞− − −
∞− − −
∞− − − −
− −
γ u γ u
γ u
/ u v /
α β u v
v
α β w
wα β w
α β α β w
x x
i i
y
πy
y
0 0
0
1
12
0
1 2
0
2
0 0
1 1 ( + )
0
1 1
0 0
1 1
0
1
0
+ 1 1 1
2 4
32
2
4 4
6 SOLUTIONS, ANSWERS, AND HINTS FOR SELECTED PROBLEMS
γ u e du u de
γu e du γ γ .
γ
γ γ
γ γ γ γ γ γ .
γ
u e du e dv ,
π
α β u v e dudv
w v v e dw dv
w v v e dv dw
w x x e dx dw ,
b x
P ω X ω x P ω X ω e e ,
G i , , i G y
y < y y <
G y
G y
G y .
. (i)
Γ( + 1) = =
= = Γ( )
(ii) An easy calculation gives Γ(1) = 0!. For an induction proof
assume that Γ( ) =
( 1)! for some positive integer . By part (i),
Γ( + 1) = Γ( ) = [( 1)!] = !
Note that the last step in the above calculation is valid for =
1. That this step be
valid is one of the motivations for the definition 0! = 1.
(iii)
Γ( ) = = 2
which, by Example 1 and symmetry, equals . Now use mathematical
induction.
(iv)
Γ( )Γ( ) =
= ( )
= ( )
= (1 )
the interchange of order of integration being valid, according
to a result from advanced
calculus, because the integrand is continuous and nonnegative.
(The validity of the
interchange in integration order is also a consequence of the
Fubini Theorem, to be
proved in Chapter 9.) The last expression is the product of the
two desired integrals.
. For = 1 and 0,
( : log ( ) ) = ( : ( ) ) = 1
the standard exponential distribution function.
. Denote the three distribution functions by , = 2 3 4. For each
, ( ) = 0
when 0 and = 1 when 2. For 0 2:
( ) = 1 1 ;
( ) = arcsin ;
( ) =
-
j
∑
2 2
1
=1
′ ′ ′ ′ ′ ′
′ ′ ′ ′ ′ ′
′ ′ ′ ′ ′ ′
4-7
4-8
4-9
4-10
4-11
4-14
s
p
s p
p Z s p s
s s
p s
ω
n n n
n n
n n
n n
n nn
nn
nn
s p g
n
j
j C
SOLUTIONS, ANSWERS, AND HINTS FOR SELECTED PROBLEMS 7
≥≤ ≤
≤
B ∞≤
b c ∧
≤
{ ≤ } { ≤ }{ ≤ ≤ }{ ≤ ≤ }
≤ { ≤ }
≤
∧ − ≤ ≤
∧ − ≤ ≤
≥ − ≥ { ∈ }
≥
≥
n p npq
/
E
E X
E X E X X
Z E X E Z E X E X
Z X E Z E X
E X E X
X X ω
, , , P P
E X X X X ω
X ω n
a b
E X E Y E X Y
E X E Y
E X X X X E Y Y Y Y
E X E Y X X,Y Y X , Y
E X Y X X,Y Y X , Y
E Z Z X Y Z E X Y .
Z Z
X Y
X n , , . . . Y n , , . . .
ω n
X ω n X ω X ω
Y ω n Y ω Y ω .
X Y Z / n Z ω ω
E X E Y E Z ,
E X E Y E X Y
E E E X
X c I .
. + (notation of Problem 39 of Chapter 3)
. 7 2
. For this problem, denote the expectation operator according to
Definition 1 by
and the expectation operator according to Definition 5 by . Let
be nonnegative
and simple. Thus, ( ) and ( ) are meaningful. Since qualifies as
an appro-
priate in the definition ( ) = sup ( ), we see that ( ) ( ). On
the
other hand, Lemma 4 implies that for all simple , ( ) ( ), from
which
it follows immediately that ( ) ( ).
. The random variable defined by ( ) = , defined on the
probability
space ((0 1] ), where denotes Lebesgue measure, has expected
value . This is
seen by calculating ( ) for simple random variables defined by (
) =
( ( ) ) .
. We treat the case = = 1. The following calculation based on
the definition
of expectation for nonnegative random variables and the
linearity of the expectation
for simple random variables shows that ( ) + ( ) ( + ):
( ) + ( )
= sup ( ) : and simple + sup ( ) : and simple
= sup ( ) + ( ): and simple
= sup ( + ): and simple
sup ( ) : + and simple = ( + )
To prove the opposite inequality, let be a simple random
variable such that
+ . By the construction given in the proof of Lemma 13 of
Chapter 2, we can find
sequences ( : = 1 2 ) and ( : = 1 2 ) of simple random variables
such
that for all and all ,
( )1
2( ) ( ) and
( )1
2( ) ( )
It is easily checked that + 1 2 for max ( ) : Ω . Thus
sup ( ) + sup ( ) ( )
and the desired inequality ( ) + ( ) ( + ) now follows from the
definition
of expected value.
. For this problem, denote the expectation operators according
to Definition 1,
Definition 5, and Definition 8 by , , and , respectively. Let be
simple (but
not necessarily nonnegative). We use (4.1):
=
-
2
∑∑
∑∑
∑
j
j
j
j
j
j
≥
−
≥
− −
−
→∞
→∞
− −
−
{ ≤ ≤ }
−
−
−
∞{ | | ∞}
− −
→ →∞
− → −
− → −
− −−
−
◦ ∞ ≤
4-21
4-22
4-23
4-26
4-29
4-30
4-31
4-35
8 SOLUTIONS, ANSWERS, AND HINTS FOR SELECTED PROBLEMS
+
: 0
: 0
+ +
: 0
: 0
+
=1
1 1
1
1 1
1 2
1 1
1
1 1
1
12 (1+ )
(1 )
2 2
+
1
1 34
2 2 443
j
j c
j C
j c <
j C
p s
j c
j j
p s
j c <
j j
g p p
n
j
j j s
n n
n
n
nn
nn
pp
p p
p
αα β
kk
ππ
C j n
X c I
X c I .
E X E X c P C
E X E X c P C .
E X E X E X c P C E X .
E X E X
P ω X ω ω
Y ω X ω X ω Y ω X ω X ω ω
Y ω Y ω
Y , Y , . . . E Y E Y n
E X X E X X .
E X
E X E X E X E X .
E X
E X E X
E X λ E X λ λ
X b b X
E X b
b E X E X b
b
,
E X /k E X k k >
E X E X E X E X
Since : 1 is a partition,
=
and
=
For these nonnegative simple random variables we have, using
Problem 9, that
( ) = ( ) = ( )
and
( ) = ( ) = ( )
By these formulas and Definition 8,
( ) = ( ) ( ) = ( ) = ( )
. The case where ( ) = + is easily treated, so we assume ( ) is
finite and,
therefore, ( : ( ) = ) = 0. Accordingly, except for belonging to
some
null set, we may define ( ) = ( ) ( ) and ( ) = ( ) ( ). For
in
the null set we set ( ) = ( ) = 0. Applying the Monotone
Convergence Theorem
to the sequence ( ), we deduce that ( ) ( ) as . It follows,
by
property (iii) of Theorem 9, that
lim ( ) ( )
Since ( ) is finite we may apply property (i) of Theorem 9 to
conclude
lim [ ( ) ( )] ( ) ( )
Now add ( ) to both sides.
. ( ) = , ( ) = (notation of Problem 11 of Chapter 3)
( ) = , ( ) = + (notation of Problem 37 of Chapter 3)
The distributions of and are identical. By Theorem 15 they have
the
same mean. By properties (i) and (ii) of Theorem 9, these equal
numbers are ( )
and ( ). It follows that ( ) = .
. (notation of Example 1 of Chapter 3)
. For standard beta distributions (that is, beta distributions
with support [0 1]),
the answer is (notation of Example 3 of Chapter 3).
. ( ) = 1 , (exp ) = if 1 and = if 1
. ( ) = ( ) = , ( ) = , ( ) =
-
Z
2
2
2− −
∞ ∞ ∞
∞ ∞
For Chapter 5
( )( )( )
∑ ∑ ∑∑ ∑
π
ππ
k
s k
k
s k
k
s k
k
s k
k
s k
5-7
5-13
5-14
5-17
5-29
5-32
− ◦ ≥ −
◦ − ◦ − ◦− ◦
≥ −
− −
−{∞}{ } ∈
−− − − − −−− − − − −
| || |
{ }
−
1 32( 8)
232 3
12 429
+
2 2 3
2 2 3
=0
2
=0
2
=0
3
169
=0
3427
=0
3
SOLUTIONS, ANSWERS, AND HINTS FOR SELECTED PROBLEMS 9
X X X X
aX b, cY d ac X,Y
ϕ
X E X ϕ X X E X ϕ E X .
X, ϕ X E X E X ϕ X E ϕ X
E X E X ϕ X
E X E X ϕ E X .
E X
n
s , s , s , s , s , s , s , s ,
s , s , s n, k k > n S , S , S ,
S , S , S , S , S , S , S ,
S n, k k > n
ρ ρ
Q Q ρ
Q k k ρ s
ρ ss s s s s
s/ s/ s/
/
s/
/
s/.
s <
s <
ρ s
ρ s
ρ s
Q k
ρ s k
k k k .
. Var( ) = Var( ) = , Var( ) = , Var( ) =
. Cov( + + ) = Cov( )
. An inequality, based on the fact that is increasing will be
useful:
[ ( )] ( ) [ ( )] ( ( ))
The following calculation then completes the proof:
Cov( ) = [ ( )][ ( )]
= [ ( )] ( )
[ ( )] ( ( )) = 0
(A slightly longer, but possibly more transparent proof,
consists of first reducing the
problem to the case where ( ) = 0 and then using the above
argument for that
special case.)
. For Example 2 of Chapter 4, the answer is 1; for Problem 18 of
Chapter 4, the
answer is 0 or 1 according as is odd or even.
. (0 0) = (1 1) = (2 2) = (3 3) = 1, (1 0) = (2 0) = (3 0) = 0,
(2 1) =
1, (3 1) = 2, (3 2) = 3, ( ) = 0 for ; (0 0) = (1 1) = (2 2)
=
(3 3) = 1, (1 0) = (2 0) = (3 0) = 0, (2 1) = (3 1) = 1, (3 2) =
3,
( ) = 0 for
. (1 ) = 1. Thus, if is the probability generating function of a
distribution
, then ( ) = 0. To both show that is a probability generating
function and
calculate ( ) for each we rewrite ( ) using partial
fractions:
( ) =24
2+
8
(2 )+
24
3+
16
(3 )+
8
(3 )
=12
1 ( 2)+
2
(1 ( 2))+
8
1 ( 3)+
16 9
(1 ( 3))+
8 27
(1 ( 3))
The first two of the last five functions are equal to their
power series for 2 and the
last three for 3. So we can expand in power series and collect
coefficients to get a
power series for ( ) that can be differentiated term-by-term to
obtain the derivatives
of ( ). Thus, we only need to show that the coefficients are
nonnegative in order to
conclude that ( ) is a probability generating function, and then
the coefficients are
the values ( ).
Formulas for the geometric series and its derivatives give
( ) = 12 ( ) + 2 ( + 1)( ) + 8 ( )
+ ( + 1)( ) + ( + 1)( + 2)( )
-
5-33
6-6
nc n n
ncn
cn
For Chapter 6
( )
(⋃ ⋂ )⋂( ⋂ )⋂ ⋃
−
′
′′
′ ′′
→∞
∞ ∞
∞ ∞
∞ ∞
→∞
k k
p
p
pm
p
nnc
n m n
m
c
n m n
m
c
n m n
cm
n
cn
A An
A
nA
nA A
{ } −
− − − −
− − − − − −
−
∞{∞}
| − | ∞
{ } −− −
{ }
− − { }
{ − } { }
10 SOLUTIONS, ANSWERS, AND HINTS FOR SELECTED PROBLEMS
Q kk k k
.
ρ s
ρ ss s s s
ρ ss s s s s s
.
s
ρ ρ .
ρ
Q
p Q
p < k m <
Q mm
m
mp p .
k k Q k
A A
A
A A .
I I I
I I I .
1
2
+3
3 3 2 4
4 3 3 4 2 5
72
72
372
744
494
254
52
=1 =
=1 =
=1 =
(lim sup ) lim sup
lim inf
When we collect terms we get nonnegative—in fact,
positive—terms, as desired:
( ) =5
2+
4 + 60 + 272
3
To get the mean and variance it seems best to work with ( ) in
the form originally
given and use the product rule to get the first and second
derivatives:
( ) =16
(2 ) (3 )+
24
(2 ) (3 )
and
( ) =48
(2 ) (3 )+
96
(2 ) (3 )+
96
(2 ) (3 )
Insertion of 1 for gives
(1) =7
2and (1) = 15
Hence, the mean equals and the second moment equals 15 + = .
Therefore, the
variance equals = and the standard deviation equals .
Had the problem only been to verify that is a probability
generating function, we
could have, while calculating the first and second derivatives,
seen that a straightfor-
ward induction proof would show that all derivatives are
positive, and an appeal to
Theorem 14 would complete the proof.
. The mean is and thus the variance is undefined. The
distribution corre-
sponding to the probability generating function with parameter
satisfies ( ) =
1 2 . Also, for 0 = 2 ,
( 2 ) =2 2 2
1[ (1 )]
For odd and = 0, ( ) = 0.
. Method 1: Using Problem 4, we get
(lim inf ) =
=
= = lim sup
Method 2: We prove that the indicator functions of the two sets
are equal:
= 1 = 1 lim sup
= lim inf (1 ) = lim inf =
-
th
=1 =
=
6-8
6-9
6-13
6-15
(⋂ ⋃ )(⋃ )
( )( )
∑ ∑∑∑ ∑⋃
∪ ∪
∩ ∩
\ \
∪ ⊇ ∪ ⊇ ∪
∩ ⊆ ∩ ⊆ ∩
\ ⊆ \ ⊆ \
4 ⊆ 4
4 ⊇ 4
∅ ⊂∅
∅ ⊂
∅ ⊂
≥
≥
− ≥ −
≤
{ }∞
∞∞
SOLUTIONS, ANSWERS, AND HINTS FOR SELECTED PROBLEMS 11
→∞ →∞ →∞
→∞ →∞ →∞
→∞ →∞ →∞
→∞ →∞ →∞ →∞ →∞
→∞ →∞ →∞ →∞ →∞
→∞ →∞ →∞ →∞ →∞
→∞ →∞ →∞
→∞ →∞ →∞
→∞
∞ ∞
→∞
∞
→∞
A B A B
A B A B
A B A B
A B A B A B
A B A B A B
A B A B A B
A B A B
A B A B .
B
n A n
A B n A
n
A B n A n
P A P A
P A P A ,
P A P A
P A P A ,
P A P A .
A
A ω X ω A A
X A
P A P A
P A P A <
P A E X P A E X
n C m , , . . . , n
m m i P C
nn n
nn
nn
nn n
nn
nn
nn n
nn
nn
nn
nn
nn n
nn
nn
nn
nn
nn n
nn
nn
nn
nn
nn n
nn
nn
nn n
nn
nn
nn n
nn
nn
n
n
ncn n
n n n
nn
n m n
m
nm n
mn
n
cn
cn
cnc
n
cnc
n
n
n n n n
n n n
n n
n n
n n n n n n
m
m
.
lim sup( ) = (lim sup ) (lim sup ) ;
lim inf( ) = (lim inf ) (lim inf ) ;
lim inf( ) = (lim inf ) (lim sup ) ;
(lim sup ) (lim inf ) lim inf( ) (lim inf ) (lim inf ) ;
(lim inf ) (lim sup ) lim sup( ) (lim sup ) (lim sup ) ;
(lim sup ) (lim sup ) lim sup( ) (lim sup ) (lim inf ) ;
lim inf( ) (lim sup ) (lim inf ) ;
lim sup( ) (lim sup ) (lim sup )
Problem 6 is relevant for this problem, especially the fifth
equality given in the problem.
Here are some examples in which the various subset relations
given above are strict.
The first and seventh subset relations above are both strict in
case = Ω for all
and = or = Ω according as is odd or even. The second, fourth,
fifth, and
eighth subset relations are all strict if = for all and = Ω or =
Ω
according as is odd or even. The third and sixth subset
relations are both strict if
= for all and = Ω or = Ω according as is odd or even.
. The middle inequality is obvious. Using the Continuity of
Measure Theorem in
Chapter 6, we have
(lim sup ) =
= lim lim sup ( )
thus establishing the first inequality. For the third
inequality, deduce from the first
inequality that (lim sup ) lim sup ( ), which is equivalent
to
1 (lim sup ) lim sup[1 ( )]
which itself is equivalent to
(lim sup ) lim inf ( )
By Problem 6, the event in the left side equals lim inf , as
desired.
. Let = : ( ) = 1 . By Problem 5, the event = lim sup is
that event that = . The events are pairwise negatively
correlated or
uncorrelated, so by the Borel-Cantelli Lemma, ( ) = 1 if ( ) = ,
and by
the Borel Lemma, ( ) = 0 if ( ) . The proof is now completed by
noting
that ( ) = ( ), so that ( ) = ( ), whether finite or
infinite.
. Let denote the number of cards and , for = 1 2 , the event
that
card is in position . The term of the formula for ( ) in Theorem
6
-
−
−
( )
( ⋃ ) ∑ ( ) ∑
∏
For Chapter 7
7-3
7-10
7-11
7-17
+1
=1 =1
+1
=1
+1
1
1 2
1
1
1 1
1
1
4
=2
i ni
n
m
m
n
i
i
n
i
i
n n n n
n
n
ncn
c cn
c
n
n
n n n n
kβ
m
ncm
cn
12 SOLUTIONS, ANSWERS, AND HINTS FOR SELECTED PROBLEMS
−
− −
− − −
− →∞
D { } ⊆ D
\ − − \
DD
D DE E
E E ⊆ EE
E E ∈ E∈ E
→ →∞ E E∈ E
∈ E∈ E
∪ → ∪ → ∞ E ∪ ∈ E∪ ∈ E
− −6
i i
n i n i
n
P Cn
i
n i
n i,
e n
A P A Q A A B
P B A P B P A Q B Q A Q B A .
A ,A , . . .
P Q
P A P A Q A Q A .
σ
A n , , . . . A,A,A, . . . A
R A n , , . . . R A ,R A ,R A , . . .
R A
R A
A A,A,A, . . .
B
B n , , . . . B
B B n B
B
B B C
C n , , . . . C
B C B C n B C n
B C
k A
A n m A A
consists of the factor ( 1) and terms each of which equals the
probability that
each of a particular cards are in a particular positions. This
probability equals the
number of ways of placing the remaining cards in the remaining
positions,
divided by !. We conclude that
= ( 1)( )!
!= ( 1)
1
!
which approaches 1 as .
. Let = : ( ) = ( ) . Suppose that are both members of .
Then
( ) = ( ) ( ) = ( ) ( ) = ( )
Thus, is closed under proper differences. Now consider an
increasing sequence
( ) of members of . By the Continuity of Measure Theorem,
applied to
both and ,
(lim ) = lim ( ) = lim ( ) = (lim )
Hence is closed under limits of increasing sequences, and
therefore is a Sierpiński
class. It contains and so, by the Sierpiński Class Theorem it
contains ( ), as desired.
. The sequences ( : = 1 2 ) and ( ) have the common limit .
By the lemma, the sequences ( ( ) : = 1 2 ) and ( ( ) ( ) ( ) )
have
equal limits. The limit of the second of these numerical
sequences is obviously ( ),
so ( ) is also the limit of the first sequence of numbers.
. Every member of is the limit of the sequence ( ). Thus .
It remains to prove that is a field.
The empty set, being a member of , is also a member of . Let .
Then
there exists a sequence ( : = 1 2 ) that converges to . By
Problem 8 of
Chapter 6, as . Since is a field, each is a member of .
Therefore
.
Let and be as in the preceding paragraph and let . There exists
a
sequence ( : = 1 2 ) that converges to . By Problem 8 of Chapter
6,
as . Since is a field, for each . Therefore
.
. The probability is 1 (1 ). The correlation between two
events
and is easily calculated; it is 0 when = . Similarly, for and .
Thus, the
Borel-Cantelli Lemma may be used to calculate the probabilities
of the limit supremum
-
R
R
−∞
−
= +1
2
2
For Chapter 8
∏
∫ ∫ ∫∫ ∫ ∫ ∫ ∫
∫
∫∫ ∫∫ ∫ ( ) ∫
7-24
7-25
7-26
7-29
8-8
ncn
n
β
k n
β
n
n n n
n n
n
n
n
n
n n
−≤
{ }
{ } − ∞
≤ ∞
L S ⊂L
− −
− ≥
− ≥ − − ≥
∞
− ≥ − ≥
−
| |∞ →| − | ≥
| − | | − |
SOLUTIONS, ANSWERS, AND HINTS FOR SELECTED PROBLEMS 13
P A P A
P A β β >
P ω Y ω
P ω Z ω n n k n <
β β > n
ϕ ϕ π
s
ϕ ϕ
r ϕ θ s r, θ
D
πr D
g f n
g f dµ g f dµ g f dµ .
gdµ <
gdµ f dµ g dµ f dµ .
g dµ
g f
f f
f dµ <
f dµ f dµ
f f n g
f f dµ f f dµ dµ .
and limit infimum.
(lim inf ) = 1 (lim sup ) = 0
(lim sup ) = 1 if 1 and = 0 if 1
( : ( ) = 1 ) = 1
( : ( ) = ) = (1 ) if
and = 1 or = 0 according as 1 or 1 if =
. Since is in one-to-one measure-preserving correspondence with
, we
only need show that the effect of a rotation or translation on
corresponds to a
transformation on having Jacobian 1, provided we identify with
+2 . It is clear
that rotations about the origin have this property, leaving
unchanged and adding a
constant to . Translations also have this property since they
leave unchanged and
add cos( ) to , where ( ) is the polar representation of the
point to which
the origin is translated.
The measure of the set of lines intersecting a line segment is
twice the length of
that line segment.
The measure of the set of lines intersecting a convex polygon is
the perimeter of
that polygon.
. The expected value, whether finite or infinite, is twice the
length of divided
by 2 . (It can be shown that this value is correct for arbitrary
curves contained in
the interior of the circle.)
. Application of the Fatou Lemma to the sequence ( : 1) of
nonnegative
measurable functions gives
lim inf ( ) lim inf( ) = ( lim sup ) 0
Since , we may use linearity to obtain
lim sup lim sup 0
Subtraction of followed by multiplication by 1 gives the last
two inequalities
in (8.2). The first two inequalities in (8.2) can be obtained in
a similar manner using
+ , and the middle inequality in (8.2) is obvious.
Under the additional hypothesis that lim = , the first and last
finite quantities
in (8.2) are equal, and therefore all four finite quantities are
equal. Thus
and . Applying what we have already proved to the sequence
( : 1), each member of which is bounded by 2 , we obtain
lim = lim = 0 = 0
-
R
2
3
3
Hint:
∫ ∫{
∣∣ ( )∣∣ ( )
8-12
8-22
8-26
− − −
→∞
−
−
− −
− − −
−
−
−
−
−
−
−
−
1 1 1
2
1 2 1 2
1 2 12
2 1 1 2
1 2 12
2 1 13
3 3 2 1 2
2
1 2
1 2 1 2
1 2
1 2
2
2
14 SOLUTIONS, ANSWERS, AND HINTS FOR SELECTED PROBLEMS
t,c t
t t,c tpt,c t
p ptp p
γγ
v /
γ/ /
γ
/ /
/ / /
γ γv
γ/
γ
/ /
/
v
/
n x
n x
{ | | ≥ }
| | | | | | ≤ | | ≤ → →∞
≥
≥
◦ − −
◦
− − −
− − −− → ∞
≥ ≤∞ ≤
−
−
≥ ≥−
√ −
| − | ≤√
I ω X ω c
E X I E X I X c E X c k c .
θ dλ θ dλ ,
λ
θ ve v
.
v
θ v γ vγ vγ .
θ v
γ vγ v γ vγ
γ vγ v γ v γ vγ ,
v / γ
θ γ θ x
v e v , θ v
vγ θ v
γ vγ vγ .
γ x vγ
xx x
x,
x
x x γ
v v
γ v e
vγ
γ
nx n
nx e ,
x x n n
x x e n
. Let denote the indicator function of : ( ) .
( ) = ( ) ( ) 0 as
. By Theorem 14 the assertion to be proved can be stated as:
lim =
where denotes Lebesgue measure on and
( ) =if 0
0 otherwise
The plan is to use the Dominated Convergence Theorem. Thus we
may restrict our
attention to 0 throughout.
We take logarithms of the integrands:
(log )( ) = ( 1) log(1 + )
The Taylor Formula with remainder (or an argument based on the
Mean-Value Theo-
rem) shows that (log )( ) lies between
( 1)( )
and
( 1)( + )
both of which approach 2 as . Thus, to complete the proof we
only need
find a dominating function having finite integral.
The integrands are nonnegative. It is enough to show, for 1,
that ( )
(1 + ) , since this last function of has finite integral on [0
). Clearly, ( )
(1 + ) ( ), the logarithm of which equals
log(1 + )(7.1)
Differentiation with respect to and writing for gives
log(1 + )(2 + )
2(1 + )(7.2)
a function which equals 0 when = 0 and is, by Problem 21, a
decreasing function of
. Thus, (7.2) is nonpositive when 0. For 1 [which we may assume
without
loss of generality], (7.1) is no larger than the value log(1 + )
it attains when
= 1. The exponential of this value is the desired function (1 +
) . [Comment:
The introduction of the factor (1 + ) in the sentence containing
(7.1) was for the
purpose of obtaining a decreasing function of .]
. The absolute value of the integral is bounded by
2 max log 1 + max
where each maximum is over those for which . Apply the
Mean-Value
Theorem to the logarithmic function, standard methods of
differential calculus to the
function , and the Stirling Formula to !. (Note: If one works
with the
-
R
For Chapter 9
k
k
k k
k
k k
k
k k
k
k
k
−
↘ ∞
∈ ∈ ≤ ∈ ∈
≤ ∈ ∈ ≤ ∈ ∈
∈ ∈
−
F G FG F R
− −
∈ E ∈≤ ≤ E
D { ∩ }
8-35
9-1
9-6
9-7
9-10
9-14
∫
∫
∫
∏ ∏
⋃(⋂ ) (⋂ ⋃ ) ( ⋃ ⋂ )
∑ (⋂ ) ∑ ∏∏∑ ∏
1 2 1 1 2
26 36 2
0 [(1 ) )
1 2 1 2
=1
=1 ( : )
( : ) ( : )
=1
SOLUTIONS, ANSWERS, AND HINTS FOR SELECTED PROBLEMS 15
n
x
A
b
a
b
a
ε n n n n n /n ,
k k k,i
k k
k
r
i
k,i
k K
k
k K
r
i
k,i
i r k K k K
k,i
i r k K k K
k,i
i r k K k K
k,i
k K
r
i
k,i
k K
k
B
x x
x e
σ ν
ν A f dλ ,
λ f ν
ν a, b f x dx
a < b
µ a, b F b F a f x dx
a < b µ ν a, b
x F x ε F F I
ε
λ λ / λ λ
B σ k K k A
i r
B A .
P B P A P A
P A P A
P A P B .
B
D P D B P D P B .
product of the maximum of the function and the maximum of the
function
one does not get an inequality that is sharp enough to give the
desired
conclusion.)
. Define a -finite measure by
( ) =
where denotes Lebesgue measure on , so that is the density of
with respect to
Lebesgue measure. In particular,
(( ]) = ( )
for all . By an appropriate version of the Fundamental Theorem
of Calculus,
(( ]) = ( ) ( ) = ( )
for all . Thus, and agree on intervals of the form ( ]. By the
Uniqueness
Theorem, they are the same measure.
. Ω and Ω each have six members, Ω has 36 members. Each of , ,
and
has 2 = 64 members. has 2 members and has 64 members.
. 1 lim [1 ( + )] and . The example = shows
that one may not just set = 0 in the first of the two
answers.
. exponential with mean ( + )
. Fix ( ) for . For each such there are disjoint members ,
1 , of such that
=
Hence,
= =
= = ( )
= ( ) = ( )
(Contrast this proof with the proof of Proposition 3.)
. For each event , let
= : ( ) = ( ) ( )
-
R
1
1 1
n
n n
Hint:
∫ (∫ )∫
∫ (∫ )∫
B
B
B A
A
k k
k k k k
B
B
A
A
1 2
1 2 2
1 2
1
1 2
2 1
2
2 1
2
23
9-15
9-23
9-27
9-29
9-33
16 SOLUTIONS, ANSWERS, AND HINTS FOR SELECTED PROBLEMS
D
E ⊆ D ∈ EE ⊆ D ∈ E E ⊆ D
∈ E E ⊆ D∈ E
∩ · · · ∩
{ ∈ } ∈ G
◦{ } { ∈ }
◦∈ G × H
C ⊆ ×C
C G × H
×
| | | | | |
| | | | | | | | ∞
∞ −∞
L B
σ B
A σ σ
A σ
A , . . . , A
P A A P A . . . P A .
x y
x f x, y B
B f y
x I f x, y
x f x, y B
f I f B
f A
x I x, y
A
A x I x, y
y
X Y X, Y
Q Q
E XY x y Q dy Q dx
x E Y Q dx E X E Y < .
E XY xy Q dy Q dx
xE Y Q dx E X E Y .
E X Y E X E Y
E X E Y
Clearly each is closed under proper differences. By continuity
of measure it is also
closed under monotone limits and, hence, it is a Sierpiński
class.
Denote the two members of by 1 and 2. By hypothesis, for each
.
By the Sierpiński Class Theorem, ( ) for each . Therefore
for
each ( ). Another application of the Sierpiński Class Theorem
gives ( )
for every ( ), which is the desired conclusion.
. The criterion is that for each finite subsequence ( ),
( ) = ( ) ( )
. Let us first confirm the appropriateness of the hint. Because
the proposition
treats and symmetrically, we only need prove the first of the
two assertions in
the proposition. To do that we need to show that : ( ) for
every
measurable in the target of and every . Suppose that we show
that the -valued
function ( )( ) is measurable. Then it will follow that the
inverse image
of 1 of this function is measurable. Since this inverse image
equals : ( ) ,
the assertion in the hint is correct.
Since is measurable, any function of the form , where is a
measurable
subset of the target of , is the indicator function of some
measurable set .
Thus, our task has become that of showing that ( ) is measurable
for each
such .
Let denote the collection of sets Ψ Θ such that ( ) is
measurable
for each fixed . This class contains all measurable rectangles,
and the class of
all measurable rectangles is closed under finite intersections.
Since differences and
monotone limits of measurable functions are measurable, the
Sierpiński Class Theorem
implies that contains the indicator functions of all sets in ,
as desired.
. The independence of and is equivalent to the distribution of (
) being
a product measure . By the Fubini Theorem,
( ) = ( ) ( )
= ( ) ( ) = ( ) ( )
Thus we may apply the Fubini Theorem again:
( ) = ( ) ( )
= ( ) ( ) = ( ) ( )
. The crux of the matter is to show that, in the presence of
independence,
the existence of ( + ) implies the existence of both ( ) and ( )
and, moreover,
it is not the case that one of ( ) and ( ) equals and the other
equals .
.
-
j
R
R
∫
∫∫ ( )
∫
√
∏
∞ −
− −
−
∞−
∞−
∞−
2 2
2 2
2 2
2 2
2 2
2 2
9-41
9-42
9-45
9-47
2
2 1 2
2
2 2
2
2
2
2
2
2
2
2
2
( )
1 2
=11
( )
→∞
−− −
√
√
√
√
≤ ≤
{ ≤ }{ }
∈≤ ≤ · · · ≤
| − |{ ≤ } ⊆ { ≤ }
x
u / σ
x / σ
x / σ
x
u / σ
x
u / σ
x
u / σ
n
i i
i
i
d d
d
d
j j/d
j j jd
d
i i
SOLUTIONS, ANSWERS, AND HINTS FOR SELECTED PROBLEMS 17
e du
σ x e.
x
e
σ x.
δ > x > σ/ δ
πσe du
<πσ
σ
ue du
<δ
πσe du .
a σ n
x v δ δ x v
j
v i x v
i x v v
j v
χ y
y y y y
d
d,
d y y y
χ
ε > x w
x w < ε
i x v i w v ε .
. Method 1: The left side divided by the right side equals
Both numerator and denominator approach 0 as ; so we use the
l’Hospital Rule.
After differentiating we multiply throughout by . The result is
that we need to
calculate the limit of1
1
The limit equals 1, as desired.
Method 2: Let 0. For ,
1
2
1
21 +
1 +
2
The expression between the two inequality signs is equal to the
right side of (9.12). (The
motivation behind these calculations is to replace the integrand
by a slightly different
integrand that has a simple antiderivative. One way to discover
such an integrand is
to try integration by parts along a few different paths, and,
then, if, for one of these
paths, the new integral is small compared with the original
integral, combine it with
the original integral. Of course, Method 1 is simple and
straightforward, but it depends
on being given the asymptotic formula in advance.)
. = 2 log
. 0
. If + for every positive , then ; hence, the infimum that one
would
naturally place in (9.13), where the minimum appears, is
attained and, therefore, the
minimum exists. As in the right side of (9.13) is increased, the
set described there
becomes smaller or stays constant and, therefore, its minimum
becomes larger or stays
constant. So (9.14) is true. The function # : has a jump of
size
# : = , possibly 0, at each . But the size of this jump equals
the number of
different values for the integer that yield this value of for
the minimum in the right
side of (9.13). Thus, (9.15) is true. The image of consists of
all for which
. For such a the cardinality of its inverse image equals
!
( !)
where denotes the number of coordinates of which equal ,
including itself.
To prove continuous it suffices to prove that each of its
coordinate functions is
uniformly continuous. Let 0. Suppose that and are members of for
which
. Then
: : +
-
√
[( ) ][ ]∑
−
−
−
−−
− ∞ − −
For Chapter 10
( )
( )
( ) ( )
1
( 1)( )
2
2
1
2
2 1
1 221
22
112 6
( )
=
9-49
9-51
9-52
9-53
9-57
10-5
10-7
10-11
10-17
i i
dj
id
j
dj
dj
d
z
ζ zζ z
zm
γ
γ
kπ
N ω n
nk n k
n λ
k λ
n k
n k k pλ
18 SOLUTIONS, ANSWERS, AND HINTS FOR SELECTED PROBLEMS
{ ≤ } ≥ ⇒ { ≤ } ≥
{ ≤ } ≥
≤
{ }
−−
∞ ≤ ∞ ≤
− − −
↘
| |
− | − | ∨
− ≤ ≤
≤ ≤
{ } { }
{ } { } −
−−
i x v j i w v ε j .
χ x v
i w χ x ε j .
χ w χ x ε x w
d ,
n , , . . .
P ω N ω nn
n.
E N e Z ,
z z e
/
E X z E X z > X < z
Xζ z ζ z ζ z
ζ zz > .
X m /m z
θγ
γθ .
γ
µ µ σ σ
x x
k
k n
P ω X ω k N ω n P ω X ω k N ω n
P ω X ω k P ω N ω nn
kp p
λ e
n.
n
pλ e
k
λ p
n k
pλ e
k,
Hence
# : = # : +
Since [ ( )] is the smallest for which the left side is true, we
have
# : [ ( )] +
Therefore, [ ( )] [ ( )] + . The roles of and may be
interchanged to
complete the proof.
. The density is ! on the set of points in [0 1] whose
coordinates are in increasing
order, and 0 elsewhere.
. For = 1 2 ,
( : ( ) = ) =( + 1)!
Also, ( ) = 1. The support of the distribution of is [0 1] and
its density there
is (1 ) .
. 1 16
. ( ) = if 2; ( ) = if 2. Var( ) = if 2 3;
Var( ) =( 2) ( ) [ ( 1)]
( )if 3
The probability that is divisible by equals 1 which approaches
as 1.
. The distribution of the polar angle has density
Γ(2 )
4 [Γ( )]sin 2
The norm is a nonnegative random variable the square of which
has a gamma distri-
bution with parameter 2 .
. normal with mean + and standard deviation +
. (1 1 ) 0
. probability at each of the points for 5 6
. For 0 ,
( : ( ) = and ( ) = ) = ( : ( ) = and ( ) = )
= ( : ( ) = ) ( : ( ) = ) = (1 )!
We sum on :
( )
!
( (1 ))
( )!=
( )
!
-
Z
Z
i
d
i
d
i
d
2
2
1
2 1
2 1
∫ ∏∫ ∫
∫ ∏ ∫
∫ ∏
10-21
10-30
−
− −
−
− −
−−
−
− −
−
− −
−
−
{ } ∈ { }
−
− 6
≥− − · · ·−
√
−
{ · · · ≤ }
−
· − −
−
· −
SOLUTIONS, ANSWERS, AND HINTS FOR SELECTED PROBLEMS 19
d
ii i
i ji j
d d
d d
d
d D
d
i
γi
i
wγ
w yγ γ
d
d
D
d
i
γi
i
wγ γ γ
d
d D
d
i
γi
i
γ γ γd
d
+
+
0
1
2
2
1 2
1 1
3 1 3 1
3 1
1 21 2
1
=3
1
02
01 2 1
11 2 3 1
2 1 1
1 21
2 1
1
=3
1
02 2
+2 3 1
2 2
1 21 2
2 1
1
=3
1+ + +1
3 1
3 1 2
Q
Q m m P m
Q δ
P P
E Yγγ , . . . , γ
Yγ γ γ
γ γ
Y Yγ γ
γ γ, i j
d d
E Y Y d
y y y d
d
D y , . . . , y y y ,
w y . . . y
E Y Yγ
γ γ γ
y
γ
y y w y y dy dy d y , . . . , y .
w y z y
E Y Yγ γ
γ γ γ
y
γy w y dy d y , . . . , y .
wz y
E Y Yγ γ γ
γ γ γ
y
γw d y , . . . , y .
γ , . . . , γ γ
as desired.
. The distribution of a single fair-coin flip is the square
convolution root. If there
were a cube convolution root , it would, by Problem 19, be
supported by . If
( ) were positive for some positive , then ( 3 ) would also be
positive,
a contradiction. Thus, it would necessarily be that is the delta
distribution , which
is certainly not a cube root of . Therefore has no cube
root.
.
( ) =1
( )
Var( ) =( )
( + 1)
Cov( ) =( + 1)
=
For the calculations of the above formulas one must avoid the
error of treating the
Dirichlet density in (10.4) as a -dimensional density on the
-dimensional hypercube.
Here are the details of the calculation of ( ) under the
assumption that 4.
We replace by 1 and discard the denominator in (10.4) in
order
to obtain a density on a ( 1)-dimensional hypercube. (In fact,
this replacement is
done so often that the result of this displacement is often
called the Dirichlet density.)
Implicitly assuming that all variables are positive, setting
= ( ): + + 1
and using the abbreviation = 1 ( + ), we obtain
( ) =Γ( )
Γ( ) Γ( ) Γ( ) Γ( )
( ) ( )
We substitute ( ) for and then use Problem 34 of Chapter 3 for
the evaluation
of the innermost integral to obtain
( ) =Γ( )
Γ( ) Γ( + + 1)
Γ( )( ) ( )
For the evaluation of the inner integral we substitute for ; we
get
( ) =Γ( )
Γ( + + + 2) Γ( )( )
By rearranging the constants appropriately we have come to the
position of needing
to calculate the integral of a Dirichlet density with parameters
, and +
-
1
− − −
Hint:
∑∑
∫ ∫
( )
d
d
d
d
j
d
i
i j
i
z
z
z z
z
w
i i i
i i
n n
n n n n
10-33
10-36
10-37
10-40
10-43
10-47
10-48
20 SOLUTIONS, ANSWERS, AND HINTS FOR SELECTED PROBLEMS
· · ·
· · ·
≤≥ ∈ −
√
− −
− −
−
−
{ ∈ ≥ }
∨ ∨ ∨ ∨
| | | | | | ∨ | |
− ≤ ≤√− −
∈ ∈ ∈
− − −
−−
∈ ∈
1
1 21 2
1
1
=1 =1
13
13
1 2 3 1 2 3
1 2 1
2 1 2
1 2 1
2 12
2 13
12
14
2 12
1 2 3
1 1 2 2 3 3 1 2 3
1 2 3 1 2 3
23
1 1 2 2
1 1 2 2 1 2 1 2
1
2 1 2
γ γ
E Y Yγ γ
γ γ.
Y Y
Y Y Y Y
F F z z
F z z z , F z /
z , z , z z > z i , , z z z
z , z z > z
z > z z z > z
F z dz dz z z .
F z z z < z <
d
,
w w w .
C C C
r x r x r x x C , r , r r r
C C C C C C
ϕ ϕ ϕ ϕ
ϕ w
wπ w w w
π.
A B
a b a b A B a A b B κ ,
κ a b κ a b κa κ a κb κ b ,
A B κa
κ a A κb κ b B A B
A B a b n
, , . . . a A b B a , b n , , . . .
+ + 2. Since the integral of the density of any probability
distribution equals 1
we obtain
( ) =( + 1)
Since + + is a constant its variance equals 0. On the other
hand, from the
formula
Var( + + ) = Cov( )
we see that the variance equals the sum of the entries of the
covariance matrix. So, in
this case, that sum is 0. But the determinant of any square
matrix whose entries sum
to 0 is 0, since a zero row is obtained by subtracting all the
other rows from it.
. Let denote the desired distribution function. Clearly, ( ) = 0
for 0 and
( ) = 1 for . Let (0 ). From (10.4), 1 ( ) equals 2 3 times the
area
of those ordered triples ( ) satisfying for = 1 2 3 and + + =
1.
This is the same as twice the area of those ordered pairs ( )
such that ,
, and 1 . Thus
1 ( ) = 2 = 1 6 + 9
Therefore ( ) = 6 9 for 0 .
. beta with parameters 1 and 2
. The distribution has support [0 ] and there the distribution
function is given
by
+ 3 + 3 log
. For , , and convex compact sets, show that
+ + : 0 + + = 1
is convex, closed, and a subset of both ( ) and ( ).
. sin , cos , sin cos
. For all and 1 1, the distribution function is
+ 1 arccos
. Let and be two compact convex sets. Consider two arbitrary
members
+ and + of + , where and . Let [0 1]. Then
( + ) + (1 )( + ) = [ + (1 ) ] + [ + (1 ) ]
which, in view of the fact that and are convex, is the sum of a
member + (1
) of and a member + (1 ) of , and thus is itself a member of +
.
Thus, convexity is proved.
It remains to prove that + is compact. Consider a sequence ( + :
=
1 2 ), where each and each . The sequence (( ) : = 1 2 )
-
2
R
Z
Z
k k
k k
For Chapter 11
({ ( ) }) ( )
( )
⋃
⋃
√
− −
−
−− −
−
−∗
−
−
∞
∞
≤
10-52
11-12
11-13
11-14
11-17
4 2 2 16
1
12
( ) 12
1+2
+
( ) 1
1 1 1
+
0
1
=0
+
n n
n n
π π π
n n
πn n
N ωk m m k
Np qr
N ωm
N
n
n n
n
m
n
m n
n
n
SOLUTIONS, ANSWERS, AND HINTS FOR SELECTED PROBLEMS 21
××
××
−
{ } { }−
− −
≤ ≤
{ − ∈ } {∞}
{∞} { ∞}
∈
{ ≤ } ∈ F
{ } ∈ F ∞
{ } ∈ F ⊆ F
{ } { ≤ } \ { } ∈ F
{ ∞} { ≤ ∞} \ { ≤ }
F{ } ∈ F ∈
∞ ∞
{ ≤ } { }
FF
a , b k , , . . . a, b A B
A B
A B a b a b A B
A B A B
ϕ
π
B
B
P ω N ω , S ω m, k rm
k mq p
m k m E S
B
P ω N ω m,S ω B Q Q B
E SQ
E S ω S ω <
N n
ω N ω n ,
n ω N ω < n <
ω N ω < n .
ω N ω n ω N ω n ω N ω < n .
ω N ω ω N ω ω N ω m
ω N ω n n
n < n
ω N ω n ω N ω m .
has a subsequence (( ) : = 1 2 ) that converges to a member ( )
of ,
because is compact. Since summation of coordinates is a
continuous function
on , the sequence ( + ) converges to the member + of + .
Hence,
+ is compact. (By bringing the product space into the argument
we have
avoided a proof involving a subsequence of a subsequence.)
. For each : mean equals and variance equals 1 +
. The one-point sets 0 and each have probability 2 3 . The
probability
of any measurable disjoint from each of these one-point sets is
the product of (1
2 3 ) and the Lebesgue measure of .
.
: ( ) 1 ( ) = ( ) =
for 2 and 0 otherwise. ( ) =
. for a Borel subset of ,
: ( ) 1 = ( ) = ( ) ( ) ;
( ) =1
( )( ; : ( ) )
. Suppose that is a stopping time. Then, for all ,
: ( )
which for = 0 is the desired conclusion : ( ) = 0 . Suppose 0
.
Then
: ( )
Therefore,
: ( ) = = : ( ) : ( )
We complete the proof in this direction by noting that
: ( ) = = : ( ) : ( )
and that all the events on the right side are members of .
For the converse we assume that : ( ) = for all . Then,
whether or = ,
: ( ) = : ( ) =
All events on the right are members of because filtrations are
increasing. Therefore,
the event on the left is a member of , as desired.
-
n
Hint:
−
−− −
−
∞
{ }
∞∗
[ ][ ]
√
( )
[ ]
∑
∑
11-24
11-28
11-30
11-32
11-40
M
n n N
M N
m m
pp
ppx
j
j
k j
k
n
ω S ω
n
n
E V
m
1
2
1
12
12
12
1 21 1
1 2
1 11
=0
: ( )=0
=0
1( )
22 SOLUTIONS, ANSWERS, AND HINTS FOR SELECTED PROBLEMS
∈ F
∩ { ≤ } ∩ { ≤ } ∩ { ≤ }
∩ { ≤ } ∩ { ≤ }
F F ∈ FF ⊆ F
− − −− ≤
{ }− { }
−− −
{ } ∞ − −≥
≥ ∞
≥
{ } { ∞ ∞ }
{ ∞} { ∞}
{ ∞}
{ }
∞ ∞∞
{ }
A
A ω N ω n A ω M ω n ω N ω n
A ω M ω n ω N ω n ,
A
σ sp p s
p s, s <
n n
n m n
m
m
mp p .
p / p
p
p p <
x
Z j
R , T , T , . . .
Z
P ω V ω k P ω Z ω , Z ω < j < k
P ω T ω P ω T ω < .
k
V P ω V ω
V I .
E V Q .
V
E V < V
P ω V ω
m /m
. Let . Then
: ( ) = : ( ) : ( )
= : ( ) : ( )
which, being the intersection of two members of , is a member of
. Hence .
Therefore .
.
( ) =1 1 4 (1 )
2(1 )0 1
For finite and even, the probability is 0 that equals the
hitting time of 1 . For
= 2 1, the hitting time of 1 equals with probability
1
2 1
2 1(1 )
The hitting time of 1 equals with probability 0 or (1 2 ) (1 )
according as
or not.
If , the global supremum equals with probability 1. If , the
global
maximum exists a.s. and is geometrically distributed; the global
maximum equals
with probability ( ) .
. Use the Stirling Formula.
. Let ( : 1) be a sequence of independent random variables with
common
distribution (as used in the theorem). From the theorem we see
that (0 )
is distributed like a random walk with steps . Thus,
( : ( ) = ) = ( : ( ) = ( ) for )
= ( : ( ) = ) ( : ( ) )
Set = 1 to obtain the first equality in (11.6). The above
calculation also shows that
is geometrically distributed unless ( : ( ) = ) = 1. Thus, it
only remains
to prove the second equality in (11.6).
Notice that
=
Take expected values of both sides to obtain
( ) = ( 0 )
If the right side equals , then = a.s., for otherwise it would
be geometrically
distributed and have finite mean. If the right side is finite,
then ( ) , and, so,
is geometrically distributed and, as for all geometrically
distributed random variables
with smallest value 1, = ( : ( ) = 1 ).
. !
-
1
2
For Chapter 12
{{{
{
[( ) ( ) ]
∑ ∑∑∫∫∫
− −
−
−
− −
− − −
−
{ | |≤ }
∞ ∞
∞
∞
∞
11-41
11-42
11-45
12-10
12-16
SOLUTIONS, ANSWERS, AND HINTS FOR SELECTED PROBLEMS 23
n n
n
n
n n nn
n n n
n
n n
n m
nk n m k n m k
m m
n
m mn
n ω X ω n n
n
n n
n n
n
m
m
m
m
c m
cm
c
c
14
( 1)
14
14
14
( 1)
12
12
12
12
1
1 1 1
1 : ( ) 1
1 1 1
1 1
1
1
=1
2
=1
=1
2 ( +1)
2
2
24
1
{∅}
{{ }} {{ }} {{ }}
{{ }} {{ }} {{ }}−
{{ }}−
{∅} −{{ }} ∞
− − −
{ − } ∞ − −
≥≥
−
| | ≤ | || | ∞ →
→
→
| |
{ | | } −
≥ −
−
−
∞ | | ∞
m Q S
Qn
n
Q Q Qn
n
Q , Q , Q ,n
n
Q , ,n
n
n
, , . . . , k n <
, . . . ,m /
n S
S n
n m
Z X I Z ω X ω n ω
E X < Z ω X ω ω X ω
E Z E X X X
Z Y
E Y E X
G X
P ω X ω > cm G cm
cG cx dx
cG cx dx
cG y dy ,
E X
. For = 3, let denote the distribution of .
( ) =(1 + 3 ) if is even
0 if is odd
( 1 ) = ( 2 ) = ( 3 ) =0 if is even
(1 + 3 ) if is odd
( 1 2 ) = ( 1 3 ) = ( 2 3 ) =(1 3 ) if is even
0 if is odd
( 1 2 3 ) =0 if is even
(1 3 ) if is odd
. probability that is hit at time or sooner: [1 ( ) ] ;
probability that
1 2 is hit at the positive time :
( ) 1 ( ) 1 ( ) ;
probability that hitting time of 1 1 equals : (2 2) (2 1)
. For 1 the distribution of assigns equal probability to each
one-point
event. The sequence is an independent sequence of random
variables. For 1, the
probability that the first return time to 0 equals is ( )(1 ) ,
where is the
number of members of the group.
. (ii) Let = . Then ( ) ( ) for each and .
Since ( ) and ( ) ( ) for every for which ( ) is finite, the
Dominated Convergence Theorem applies to give ( ) ( ). Since
and
have identical distributions, and also have identical
distributions and hence the
same expected value. Therefore ( ) ( ).
. Let denote the distribution function of . Then
( : ( ) 2 ) = [1 (2 )]
1
2[1 (2 )]
=1
2[1 (2 )]
=1
4[1 ( )]
which, by Corollary 20 of Chapter 4, equals , since ( ) = . By
the Borel-
Cantelli Lemma, (12.1) is true.
-
Z
th
12-17
12-19
12-27
−
−
−
→∞
→∞− − −
− −
−
∞
∞
∣∣ ∣∣ ∣∣ ∣∣
∏ ∑∫
⊗∏
( ⊗ )
[ ] [ ] [ ]
m m m
m m
nc
nn
k kn
nn
k k
n
n
n
ε nn
ε n
n n
nn n
n
n n
n
n
p
n
n p
| | | | | |
∨ −
→ ∞
−
{ }
− −
−−
− − −− −
∈ G
4
×
\ { }
≤− ≤
∩ ⊆ 4 ∪ ∩ ∪ 4
2 2 2 1
2 2 1
2
=1
=1
1
1
0
(1+ ) (1 )
1
2
2
2
2 3 4
2 3 4
=1
=1
= +1
+
24 SOLUTIONS, ANSWERS, AND HINTS FOR SELECTED PROBLEMS
X ω > cm S ω > cm S ω >
cm
S ω
m
S ω
m>c.
ω
m ω S ω /n
c
E S E X
S X
S
nE X xdx
ε >
P ω e < S < e .
E S /S n
S e
E S
S S
n
n
, S
p p p p
p
p p,
p p p p p
p p p
A ε > A
p D P A B < ε
B D .
π
π n
n p n p
n p p < n p
n p < n .
π
A π A A B B π B π B π A .
To prove (12.2) we note that if ( ) 2 , then ( ) or ( )
from which it follows that
( )
2
( )
2 1 2
From (12.1) we see that, for almost every , this inequality
happens for infinitely many
. Hence, 0 is the probability of the event consisting of those
for which ( )
converges to a number having absolute value less than . Now let
through a
countable sequence to conclude that (12.2) is true.
. ( ) = ( ) = 2 . An application of the Strong Law of Large
Numbers to the sequence defined by log = log gives
limlog
= (log ) = log = 1 a.s.
Since almost sure convergence implies convergence in
probability, we conclude that, for
any 0,
lim ( : ) = 1
Thus, with high probability ( ) is very large for large . There
is some small
probability that is not only much larger than , but even much
larger than 2 ,
and it is the contribution of this small probability to the
expected value that makes
( ) much larger (in the sense of quotients, not differences)
than the typical values of
. The random variable represents the length of the stick that
has been obtained
by starting with a stick of length 1 and breaking off pieces
from the stick, the length
of the piece kept (or the piece broken off) at the stage being
uniformly distributed
on (0 ).
. (1 + )(1 ), (1 + )(1 ) ,
(1 )
1 +
(1 + + )(1 )
1 + 2
. Let and 0. (We are only interested in exchangeable but the
first part of the argument does not use exchangeability.) By
Lemma 18 of Chapter 9,
there exists an integer and a measurable subset of Ψ such that (
) ,
where
= Ψ
Define a permutation of 0 by
( ) =
+ if
if 2
if 2
Let ˆ denote the corresponding permutation of Ω.
It is easy to check the following set-theoretic relation:
ˆ( ) ˆ( ) ˆ( ) ˆ( )
-
2
∞=1
2
2 2 2
2
2 2
2
2
1
2
2
12-30
12-35
12-40
n
n
n
n
n
nn
n n n n
n n n n
n
( ) ( ) ( ) ( )⊗
( ) ( ) ( )( ) ( )
( )
∑ ∑∑
∑
∩ ≤ 4 ∩ 4
∈ G
4 4 4
∩ ∩
∩
∩
≤
−
↘ −
{ | | } ≤ ∞| | ≤
{ } ≤
≤ ≤ ≤ ≤
≤ ≤ ≤
SOLUTIONS, ANSWERS, AND HINTS FOR SELECTED PROBLEMS 25
P A π A P A B P B π B P π B π A .
ε P C P π C
C
P π B π A P π B A P B A < ε .
ε
P A π A < P B π B ε
A A π A A B
π B
P B π B P B P π B P B .
P B < P A ε
P A < P A ε ε P A ε ε .
P A P A < ε ε .
ε P A P A
σ σ
σ
σ
P ω X ω > /n /n <
ω X ω /n n
X ω ω
b Y
P ω X ω > bE X
b bn.
Y X E Y
Y E Y bE Y bE Xb
n.
X
Hence
ˆ( ) + ˆ( ) + ˆ( ) ˆ( )(7.3)
The first term on the right side of (7.3) is less than . Since (
) = (ˆ( )) for any
,
ˆ( ) ˆ( ) = ˆ( ) =
Thus the third term on the right side of (7.3) is also less than
. Therefore
ˆ( ) ˆ( ) + 2(7.4)
Now assume that is exchangeable. Then ˆ( ) = . Also, it is clear
that
and ˆ( ) are independent, and so
( ˆ( )) = ( ) (ˆ( )) = [ ( )]
Another easily obtained fact is that ( ) ( ) + . From (7.4), we
therefore obtain
( ) ( ) + + 2 [ ( )] + 4 +
Algebraic manipulations give
( )[1 ( )] 4 +
Let 0 to obtain ( )[1 ( )] = 0, as desired.
. (i) exchangeable but not tail, (ii) exchangeable and tail,
(iii) neither exchange-
able nor tail (but the Hewitt-Savage 0-1 Law can still be used
to prove that the given
event has probability 0 or 1) [Comment: the tail -field is a
sub- -field of the ex-
changeable -field, so there is no random-walk example of an
event that is tail but not
exchangeable. This observation does not mean that the Kolmogorov
0-1 Law is a corol-
lary of the Hewitt-Savage 0-1 Law, because there are settings
where the Kolmogorov
0-1 Law applies and it is not even meaningful to speak of the
exchangeable -field.]
. ( : ( ) 1 ) (1 ) . By the Borel Lemma, for al-
most every , ( ) (1 ) for all but finitely many . By the
comparison test for
numerical series, ( ) converges (in fact, absolutely) for such
.
. by the Three-Series Theorem: Let be any positive number, and
define as
in the theorem. By the Markov Inequality,
( : ( ) )( )
=1
Thus the series (12.8) converges. Since 0 , 0 ( ) . Hence,
the
series (12.9) converges. Also,
Var( ) ( ) ( ) ( ) =
Thus the series (12.10) converges. Therefore, converges a.s.
(Notice that this
proof did not use the fact that the random variables are
geometrically distributed.)
-
x
2
2
R R
R
Z
Hint:
∑ ∑∑∑
∑ ∑ ∑
∑ ∑∫
( )
−
− | | −−
− −
∞−
′′
| |
− −
For Chapter 13
For Chapter 14
∞
∞
6 6
∞
→
√−
−√
−
D∈
D ↘ ↗∞
{ } → ∈− { }
n
n n
n
n
n
k km
k kn
k m k
d
pp
k pp p v
m
k
m
k
αy
n
k
k k
nλx
λ λ
λnn x
n
2 1
2 4 2
2
=1 =1 = +1+
11+
(1 )1+ 2 cos
=11
=12
02 2
2
2 2
2 2
1 2
!+
26 SOLUTIONS, ANSWERS, AND HINTS FOR SELECTED PROBLEMS
12-41
12-45
13-15
13-19
13-30
13-34
13-48
13-72
14-2
14-4
X n
n /n < /n
E X E X <
X
c <
X ω X ω X ω
k p v p
k k
f αu y
e dy
f u f
π
a b
a a b
b
x F G F x G x
F F
G F G
y
x , x , . . . x y k F
G F x G x k F y G y
Q x e x e
λ Q x
by Corollary 26: The distribution of is geometric with parameter
( + 1) .
Thus the variance is ( + 1) 2 . The series of these terms
converges, as does
the series of expectations. An appeal to Corollary 26 finishes
the proof.
by Monotone Convergence Theorem: ( ) = ( ) . A random
variable with finite expectation is finite a.s. Therefore, is
finite a.s. (Notice that
for this proof, as for the proof by the Three-Series Theorem,
the geometric nature of
the distributions was not used.)
.
. One place it breaks down is very early in the proof where the
statement
( ) = ( ) is replaced by the statement ( ) = 0. These
two statements are equivalent if the state space is , but if the
state space is it is
possible for the first of these two statements to be false, with
both sums equal to ,
and the second to be true.
. if and only if the supports of the two uniform distributions
have the same
length
. ; . [ is the parameter of the (unsymmetrized)
geometric distribution.]
. mean equals and variance equals
. Let
( ) =1
+
and find a simple formula for + .
.
2
. yes
. At any where both and are continuous, ( ) = ( ). The set of
points
where is discontinuous is countable because is monotone. The
same is true for
. The set of points where both and are continuous, and thus
equal, is dense,
because it has a countable complement. For any , there exists a
decreasing
sequence ( ) in such that as . The right continuity of and
and the equality ( ) = ( ) for each then yield ( ) = ( ).
. We will first show that for each . The factor arises
as the limit of (1 ) . The factor already appears in the formula
for , and
-
R
14-6
14-10
− −
−
∈
−
∞∨
≥
−
−
→∞
1
=0
00
1
0 0
1
1
1
00
1
=0 0
1
=1 0
1
1
0
( )
∏( )
∑ ∑
∫ [ ][ ∫ ]∫∫
∑( ) ( ) ∫
∑( ) ( ) ∫
∫
→∞
−
−
→∞≤ ≤
−
↘− −
↘ ↘
∞− −
∞ − −
− −
↘− −
→∞
b c ↑− −
→∞
b c ↑− −
SOLUTIONS, ANSWERS, AND HINTS FOR SELECTED PROBLEMS 27
n x
x
x
k
nx y
n
x y
xλ
γ
xγ u
γ γx
γ u
x
u
u
γ
xγ u
m
mx
k
r k kx
r u
m
mx
k
r k kx
r u
m
x
m
x
n
n x n
λ
n.
k
n
Q xλ
xe
y
x >
γu e du
γu e du .
u e du <
u e
γu e du
x >
x
x r >
m
r
k m ru e du ,
m
r
k m ru e du ,
k m
g u du ,
! appears there implicitly as part of the binomial coefficient.
To finish this part of the
proof we need to show
lim!
( )!1 = 1
The second factor obviously has the limit 1 and the first factor
can be written as
1
which also has limit 1.
We will finish the proof by showing that
lim ( ) =!
for every . On the left side the limit and summation can be
interchanged because
the summation has only finitely many nonzero terms. The desired
equality then follows
from the preceding paragraph.
This problem could also be done by using Proposition 8 which
appears later in
Chapter 14.
. standard gamma distributions. For 0,
lim1
Γ( )= 1 lim
1
Γ( )lim
The first limit in the product of two limits equals 0 and by the
Dominated Convergence
Theorem, the second limit equals , a dominating function
being
( 1) . We conclude that
lim1
Γ( )= 1
for 0 from which convergence to the delta distribution at 0
follows (despite the
fact that we did not obtain convergence to 1 at = 0).
. Fix 0 and 0. We want to show
lim1 ( )
!1
1=
1
Γ( )
which is equivalent to
lim1 ( )
!1
1=
1
Γ( )(7.5)
because the term , obtained by setting = 0, approaches 0 as
.
The sum on the left side of (7.5) can be written as
( )
-
kr
↑
−1
12
12
{( ) ( )
∫
( ) ( )( )
( ) ( ) ( )
( ) ( ) ( )( )
−
− −
− − − −
− −
− −
−
→∞ −
→∞
− −
− −
− −
→∞
− −
−
− −
− − ≤ b c
→ → ∞
→ ∞ → ∞∈
→ ∈ ≤ ≤≤
≤
−
√√
− −
− − ≤ −
− →
−
1 ( )
!1
0
1
1 1 1 1
1 1
1 1
1
1
+ ( + )
1 + ( +1)
( 1)1 1 2 +1
+1
1 1
28 SOLUTIONS, ANSWERS, AND HINTS FOR SELECTED PROBLEMS
m
km
r r
k k m
k
x
r u
m
m
m
km
r r km
r r
km
r r
r r
m
r
k r
k
r k r k
r k k
r
k
r / k
m
m
mu k mu
k u
mm u
m
g uk < mu k k , , . . . , mx
g u du ,
g uru e .
g u g u m u , x
h g
g
k u m k m
u , x
u u , x r u
r > x
h u r x
r >
g u
r
r k
k k.
r
r k
k k
r
π r k e
k π k e
e
r
r
k
r
k
r
k
r.
g u k
r
h
g u
m<
m m,
me .
< e e
g u
where
( ) =1 if 1 for = 1 2
0 otherwise ;
and the right side can be written as
( )
where
( ) =1
Γ( )
The plan is to show that ( ) ( ) as for each in the interval (0
)
and to find a function that has finite integral and dominates
each , for then the
desired conclusion will follow immediately from the Dominated
Convergence Theorem.
We will consider the three factors in separately. It is
important to keep in mind
that depends on and and that in particular, as for each
fixed
(0 ), as this dependence is not explicit in the notation.
It is clear that for (0 ). In case 1, . In
case 1, . Thus, we have constructed one factor of what we
hope
will be the dominating function : in case 1 and the constant in
case
1.
The second factor in ( ) equals
1
Γ( )
Γ( + )
Γ( 1)
We use the Stirling Formula to obtain the limit:
1
Γ( )lim
Γ( + )
Γ( + 1)
=1
Γ( )lim
2 ( + )
2 ( + 1)
=Γ( )
lim 1 + 1 +1
+ 11 +
1
+ 1
=1
Γ( )
The second factor in ( ) is thus bounded as a function of , the
bound possibly
depending on . Such a constant bound will be the second factor
we will use in con-
structing the dominating function .
For the third factor in ( ) we observe that
11
11
11
(7.6)
from which it follows that
11
Moreover, (7.6) and the inequality (1 ) imply that is a
dominating
function for the third factor in ( ).
-
R
∑
− −( + )ax b ax
− − −
→∞
− −
−
∞
→∞
D
↘
1
=1
1 2 1 2
1
2 1 2
1 2 1
1
14-14
14-16
14-22
14-35
14-37
r u u
p,r m m
e ce
b
n
n
n
nn
k
n
n
n
α α α
α
α α
n n n n
zz
≤
∈
∞
→∞ →∞
{ − }
−
− −→ →∞
→ →∞
− −−
− − −
{ } − { }
−∞ −
SOLUTIONS, ANSWERS, AND HINTS FOR SELECTED PROBLEMS 29
h u
u e r e r >
, x
r Q R
G
a > b
G ax b e e ,
c e >
x
P X > c .
M n
ω M ω n > m
X k , , . . . m
M n
M n
nn
M
nn .
α >
α >
Q Q n
Q , xc
c
Our candidate for a dominating function ( ) having finite
integral is a constant
multiple of in case 1 and a constant multiple of in case 1.
Both
these function have finite integral on the interval [0 ], as
desired.
For = 0, each is the delta distribution at 0, and, therefore,
lim
equals this delta distribution.
. Let denote the standard Gumbel distribution function defined
in Problem 13.
For 0 and ,
( + ) = =
where = 0.
. For any real constant ,
[ ] =
By the Borel-Cantelli Lemma, a.s. as and, hence,
: lim [ ( ) log ] exists and
is a tail event of the sequence ( : = 1 2 ) for every . By the
Kolmogorov 0-1
Law, the almost sure limit of ( log ) must equal a constant if
it exists. On the
other hand, by the preceding problem the almost sure limit, if
it exists, must have a
Gumbel distribution. Therefore, the almost sure limit does not
exist.
The sequence does not converge in probability, for if it did,
there would be a sub-
sequence that converges almost surely and the argument of the
preceding paragraph
would show that the distribution of the limit would have to be a
delta distribution
rather than a Gumbel distribution.
The preceding problem does imply that
log
log0 as
and, therefore, that
log1 in probability as
In Example 6 of Chapter 9 the stronger conclusion of almost sure
convergence was
obtained using calculations not needed for either this or the
preceding problem.
. Weibull: mean = Γ(1+ ), variance = Γ(1+ ) [Γ(1+ )] ; Fréchet:
mean
is finite if and only if 1 in which case it equals Γ(1 ),
variance is finite if and
only if 2 in which case it equals Γ(1 ) [Γ(1 )]
. 0 = 1 , =
. We need to show
lim ( ] =1
-
−
/ z1 ( 1)
R
R R
R R
R R
↘
b c
− −
−
∞−
14-44
14-48
∑
∫ ∑ ∫
( ) ∑ ( )
( ) [ ( )]
( ) ( )[ ( )] [ ( )]
∑ ( )
1=1
1( )
1 =1 1
1
=1
1
1 ( 1)
1
1 1 1 2 1 3
2 1 2 2 2 3
3 1 3 2 3 3
1 2 3
=0
z
c x
k
z
ζ z
m
z
m
k
z
m
z
z
m
k
z z
/ z
c
n n
p,r p,r
p,r
x
r x x r r
p,r
p,r
30 SOLUTIONS, ANSWERS, AND HINTS FOR SELECTED PROBLEMS
−
−↘
− − − −
b c − ↘−
| | ≤ − →
− ≤ −
− − −− − −
− − − ≥
− −
− ≥
− ≥ −
≤ 6
− − − −
{ ≤ ≥ }
x
ζ z k
c
c.
z
z
xdz <
k<
xdz
z m<
k<
z m
m c x z z
β u u n β u
u
β u β u ,
β
β
β z z β u z z β u z z
β u z z β u u z z β u u z z
β u z z β u u z z β u u z z .
z z z β v β v β
β u β u ,
β u β u ,
< u
ρ Q
ρ s pr
xp s p ps .
p, r ρ s
p, r p < , r
s p, r Q
for all positive